1. Use Carius method data to find mass of chlorine in the sample

   The chlorine in compound $A$ is converted into $AgCl$.

   Given:

   • Mass of $AgCl = 0.3849\,\text{g}$
   • Atomic mass of $Ag = 107.87$
   • Atomic mass of $Cl = 35.5$

   So, molar mass of $AgCl$ is $107.87 + 35.5 = 143.37$

2. Mass fraction of chlorine in $AgCl$

   $\text{Mass fraction of Cl in } AgCl = \frac{35.5}{143.37}$

   Therefore, mass of chlorine in $0.3849\,\text{g}$ of $AgCl$ is $m_{Cl} = 0.3849 \times \frac{35.5}{143.37}$

   $m_{Cl} \approx 0.0953\,\text{g}$

3. Find percentage of chlorine in compound $A$

   The mass of compound $A$ taken is $0.5\,\text{g}$.

   Hence, $\%Cl = \frac{0.0953}{0.5} \times 100$

   $\%Cl \approx 19.06\%$

4. Round off to nearest integer

   $\boxed{19}$

5. About the given 5.0 g chlorine

   The statement that compound $A$ is obtained using $5.0\,\text{g}$ chlorine is not needed for calculating percentage of chlorine from the Carius method data. The percentage is determined directly from the $AgCl$ formed.

Therefore, the percentage of chlorine in compound $A$ is $\boxed{19\%}$.